A = Qinside. E d. Today: fundamentals of how currents generate magnetic fields 10/7/15 2 LECTURE 14. Our Study of Magnetism

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1 LECTUE 4 Fundamental Laws for Calculating B-field Biot-Savart Law ( brute force Ampere s Law ( high symmetry Example: B-field of an nfinite Straight Wire from Biot-Savart Law from Ampere s Law Other examples Our Study of Magnetism Lorentz Force Equation Motion in a uniform B-field Forces on charges moving in wires Magnetic dipole N S Today: fundamentals of how currents generate magnetic fields /7/5 Calculation of Electric Field Two ways to calculate Calculation of Magnetic Field Two Ways to calculate Coulomb s Law E = 4πε q r ˆr "Brute force" Biot-Savart Law Gauss Law ε E d A = Qinside "High symmetry" What are the analogous equations for the Magnetic Field? "Brute force" Ampere s Law AMPEAN LOOP NTEGAL "High symmetry also, only the ENCLOSED Current These are the analogous equations

2 Biot-Savart Law: B-field due to a current in a wire db = µ o ˆr The magnetic field curls or loops around the wire No db field at P since d is parallel to r Observations about Biot-Savart Law db = µ o ˆr. rˆ points from d to the field point.. db d and db rˆ. d rˆ = d sin(d,rˆ. 4. For a given current, db is maximum when d rˆ, and db = when d rˆ. Note: A moving charge produces a B field. B = µ oq v ˆr /7/5 6 Observations about Biot-Savart ight Hand ules db = µ o ˆr Oersted s Experiment DEMO 6B- *Grasp element in your right hand with thumb pointing in the direction of the current. Your fingers naturally curl around in the direction of the magnetic field. *To find db at point P, r /7/5 7 Switch open: = Compass points north. Switch closed: B d /7/5 8

3 Magnetic Field of an nfinite Straight Wire Calculate field at point P using Biot-Savart Law: x φ φ r B = µ o π db = µ o sinϕ P y s x + B = db = µ o π r = s + sinϕ = sin(π ϕ = ( / B = µ o π ds ( s + / dx ( x + a = x / a x + a = µ o s π x + ( / ( / sinϕ r s + = µ o π Magnetic Field of an nfinite Straight Line Calculate field at distance from wire using Ampere's Law: B d l = µ o enc Ampere's Law simplifies the calculation thanks to symmetry around the current! (axial/cylindrical B Magnetic Field Lines of a Straight Wire Magnetic Field at the Center of a Current Loop Practice both right hand rules here: db = µ o 4π DEMO 6B- /7/5 /7/5

4 Magnetic Field at the center of curvature for a partial loop y d = d Only if φ in radians db = µ o 4π = µ o dφ 4π x Line Segment Combinations db = µ d ˆr db = µ d sinθ Superposition principle: B = B + B + B P Find B at point P. / is just the fraction of a full circle that carries current. Note that as usual, the current comes in and leaves via un-shown wires! /7/5 /7/5 4 Double Arc Magnetic Field on Axis of Current Loop db = µ o ˆr Find B at point P. Only the arcs contribute. /7/5 5 /7/5 6 4

5 Magnetic Field Lines of a Current Loop Solenoid (DEMO DEMO 6B-4 & 5 DEMO 6B-4 & 5 /7/5 7 Solenoid /7/5 8 Solenoid esult for B on the axis of a current loop: µ (x + n=n d d µ di (x + Bx at the end of a long solenoid is exactly half of the value below, which is for places far from either end of a long solenoid. for a solenoid = no. of turns/unit length, di = n dx L µ n dx (x + µ n x x This makes sense, since putting two such long equal solenoids end to end gets you back to full strength, and so each of the halves must contribute Bx/ dx (x + x x µn x + x + f -x >> and x >> (long solenoid, /7/5 B x = µ n 9 /7/5 emember n is the number of turns per meter. 5